If the random variable X follows the Poisson distribution such that \(P[X = k] = P[X = k+1]\), then the mean value of X is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(k + 1\)
For a Poisson distribution with parameter \(\lambda\), \(P[X = r] = \frac{e^{-\lambda}\lambda^r}{r!}\), and the mean is \(\lambda\).
\( \displaystyle \frac{e^{-\lambda}\lambda^{k}}{k!} = \frac{e^{-\lambda}\lambda^{k+1}}{(k+1)!} \)
Cancelling \(e^{-\lambda}\lambda^{k}\) and using \((k+1)! = (k+1)\,k!\):
\( \displaystyle 1 = \frac{\lambda}{k+1} \)
\( \lambda = k + 1 \)
So the mean of X is \(k + 1\).